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3 years ago

13

In the late 1 800s , experiments using cathode ray tubes led to the discovery of the

1 answer:

Mice21 [21]3 years ago

7 0

<em>In the late 1800's, experiments using cathode ray tubes led to the discovery of the;</em>

Electron(s)

<u>It was founded that many different metals release cathode rays, and that cathode rays were made of electrons, very small negatively charged particles.</u>

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What does D represent?

alexandr1967 [171]

Answer:

the lowest point of energy the the graph reaches

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3 years ago

When 8.00x10^22 molecules of ammonia react with 7.00x10^22 molecules of oxygen according to the chemical equation shown below, h

Pavel [41]

NH₃:

N = 8*10²²
NA = 6.02*10²³

n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol

O₂:

N=7*10²²
NA = 6.02*10²³

n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol

4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant

N₂:

n = 0.0665mol
M = 28g/mol

m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>

6 0

3 years ago

When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H

algol [13]

Answer: The empirical formula is $CH_2$ and molecular formula is $C_4H_8$

Explanation:

We are given:

Mass of $CO_2$ = 18.95 g

Mass of $H_2O$ = 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, = $\frac{12}{44}\times 18.59=5.07g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, = $\frac{2}{18}\times 7.759=0.862g$ of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.422}{0.422}=1$

For H = $\frac{0.862}{0.422}=2$

The ratio of C : H = 1: 2

Hence the empirical formula is $CH_2$ .

The empirical weight of $CH_2$ = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4$

The molecular formula will be= $4\times CH_2=C_4H_8$

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3 years ago

If not managed properly, aquaculture can

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C. pollute water and damage aquatic ecosystems

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3 years ago

What can one conclude from the fact that water has a higher specific heat than sand?

miss Akunina [59]

Sand will burn faster

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3 years ago